Question

A force $\overrightarrow{F}=-K\left(y\hat{\text{i}}+x\hat{\text{j}}\right)$ (where $K$ is a positive constant) acts on a particle moving in the $x-y$ plane. Starting from the origin, the particle is taken along the positive $x$ -axis to the point ( $a,0$ ) and then parallel to the $y$ -axis to the $(a,a)$ . The total work done by the force $\overrightarrow{F}$ on the particle is

• A. $-2K{a}^2$
• B. $2K{a}^2$
• C. $-K{a}^2$
• D. $K{a}^2$

Answer 1

Answer: (C)

$\text{F}=-\text{K}\left(\text{y}\hat{\text{i}}+\text{x}\hat{\text{j}}\right)$
On moving from origin to (a, 0).
$\text{Work done}={\int }_0^{\text{a}}-\text{K}\left(\text{y}\hat{\text{i}}+\text{x}\hat{\text{j}}\right)\text{.}\text{dx}$
= 0
as y = 0
Work done from (a, 0) to (a, 0)
Work done $\text{w}={\int }_0^{\text{a}}-\text{K}\left(\text{y}\hat{\text{i}}+\text{x}\hat{\text{j}}\right)\text{dy}$ but x = a
$={\int }_0^{\text{a}}-\text{K}\left(\text{a}\right)\text{dy}$
$=-{\text{Ka}}^2$